Circular Motion with Decreasing Radius

AI Thread Summary
The discussion centers on the relationship between tangential speed and radius in circular motion. Initially, there is a claim that the new tangential speed should double when the radius is decreased by a factor of four, but this is proven incorrect. The correct interpretation shows that if the radius decreases, the tangential speed must actually halve to maintain the centripetal force. The participants clarify the equations involved, ultimately confirming that the new speed is indeed half of the original speed. The misunderstanding is resolved, highlighting the importance of accurately applying the principles of centripetal force.
uSee2
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Homework Statement
One end of a string is attached to a ball, with the other end held by a student such that the ball is swung in a horizontal circular path of radius R at a constant tangential speed. At a later time, the tension force exerted on the ball remains constant, but the length of the string is decreased to R/4. What is the new tangential speed of the ball?
Relevant Equations
##F_c = \frac {mv^2} R##
The answer key states that the new tangential speed is half the original speed. However, this isn't correct right? It should double.

My proof:

##F_c = \frac {mv^2} R##
##F_c = F_t##
##\frac {mv^2} {\frac R 4} = \frac {m(2v)^2} R## If centripetal force were to stay constant.
As such, tangential velocity should double right? It should not half?
 
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Your logic is faulty. The centripetal force is, as you say, ##F_c =\frac{mv^2}{R}.## For the fraction to remain constant, if the denominator is decreased by a factor of 4, the numerator must also decrease by a factor of 4. Since the speed in the numerator is squared (##v^2=v*v##), each of the v's must decrease by a factor of 2 when the radius is decreased by a factor of 4.
 
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kuruman said:
Your logic is faulty. The centripetal force is, as you say, ##F_c =\frac{mv^2}{R}.## For the fraction to remain constant, if the denominator is decreased by a factor of 4, the numerator must also decrease by a factor of 4. Since the speed in the numerator is squared (##v^2=v*v##) each of the v's must decrease by a factor of 2 when the radius is decreased by a factor of 4.
But couldn't the denominator decrease and the numerator increase instead and give the same ratio?

##\frac {mv^2} {\frac R 4} * \frac 4 4 = \frac {4mv^2} R##
Then it is equal to
##\frac {m(2v)^2} R##
Isn't this true?
 
uSee2 said:
But couldn't the denominator decrease and the numerator increase instead and give the same ratio?

##\frac {mv^2} {\frac R 4} * \frac 4 4 = \frac {4mv^2} R##
Then it is equal to
##\frac {m(2v)^2} R##
Isn't this true?
Try using two different symbols for the first velocity and the second velocity.

$$\frac {mv_1^2} {R} = \frac {mv_2^2} {\frac R 4}$$

Well I cannot get my Latex to display correctly, but try it on your own.
 
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uSee2 said:
But couldn't the denominator decrease and the numerator increase instead and give the same ratio?
No. The new ratio must be the same as the old ratio.
Old ratio = ##\frac{mv^2}{R}##.
New ratio (according to you) = ##\frac{m(2v)^2}{R}.##

Are they the same?
 
uSee2 said:
But couldn't the denominator decrease and the numerator increase instead and give the same ratio?

##\frac {mv^2} {\frac R 4} * \frac 4 4 = \frac {4mv^2} R##
Then it is equal to
##\frac {m(2v)^2} R##
Isn't this true?
Yes, and that is exactly what the answer key says. The new tangential speed, v, is half the original, 2v.
 
haruspex said:
Yes, and that is exactly what the answer key says. The new tangential speed, v, is half the original, 2v.
kuruman said:
No. The new ratio must be the same as the old ratio.
Old ratio = ##\frac{mv^2}{R}##.
New ratio (according to you) = ##\frac{m(2v)^2}{R}.##

Are they the same?
I do see that they aren't the same. But when I tried it again with different symbols just like @scottdave said:

2 is after the label for when radius is reduced, 1 is the label for before.

##F_{c2} = \frac {mv_2^2} {R_2}##

##F_{c1} = \frac {mv_1^2} {R_1}##

##F_{c1} = F_{c2}##

##R_2 = \frac 1 4 {R_1}##

##\frac {mv_1^2} {R_1}= \frac {mv_2^2} {R_2}##

##\frac {mv_1^2} {R_1}= \frac {mv_2^2} {\frac 1 4 R_1}## Substitution of ##R_2 = \frac 1 4 {R_1}##

Multiply each side by ##\frac {R_1} {m}##

##\frac {v_2^2} {\frac 1 4} = v_1^2##

##v_2^2 = 4v_1^2##

Take the square root of each side

##v_2 = 2v_1##

The new speed is double the original speed I think, I may have made an error though.
 
uSee2 said:
##\frac {v_2^2} {\frac 1 4} = v_1^2##

##v_2^2 = 4v_1^2##
Check how you went from the first equation to the second equation.
 
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TSny said:
Check how you went from the first equation to the second equation.
Ohh!! I can't believe I made that mistake. Thank you so much for catching that. I see now that ##v_2 = \frac 1 2 v_1##
 
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